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I wanted to see how much calculus I could derive from the ground up using basic definitions.

I was able to show that for example

\begin{align} \frac{d}{dx}cx^n &= \lim_{h \to 0} \frac{c \cdot(x+h)^n - cx^n}{h} \\&= \lim_{h \to 0} \frac{c}{h} \cdot((x+h)^n - x^n) \\&= \lim_{h \to 0} \frac{c}{h} \cdot\left(-x^n + \sum_{k=0}^{n}\binom{n}{k}x^kh^{n-k}\right) \\&= \lim_{h \to 0} \frac{c}{h} \cdot\left(\sum_{k=0}^{n-1}\binom{n}{k}x^kh^{n-k}\right) \\&= \lim_{h \to 0} \frac{c}{h} \cdot\left(nh\sum_{k=0}^{n-1}\binom{n-1}{k}x^kh^{n-1-k}\right) \\&= \lim_{h \to 0} \frac{c}{h} \cdot\left(nh(x+h)^{n-1}\right) \\&= \lim_{h \to 0} cn(x+h)^{n-1} \\&= cnx^{n-1} \end{align}

Which surprised me that it actually worked!

However when I tried the same thing for chain rule I was hopelessly stuck:

\begin{align} \frac{d}{dx}f(g(x))&= \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h} \end{align}

And sadly that is as far as I could go! I have no idea how you're supposed to simplify that any further.

Or should it be

\begin{align} \frac{d}{dx}f(g(x))&= \lim_{h \to 0} \frac{f(g(x)+h) - f(g(x))}{h} \end{align}

instead?

Simply Beautiful Art
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user51819
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  • The first fraction (with $f(g(x+h))$ in the numerator) is the correct one. – Arthur Sep 09 '16 at 20:53
  • According to the definition of derivative, the derivative of $f(y)$ is equal to $\frac{d}{dy}f(y) = \lim_{h \to 0} \frac{f(y+h) - f(y)}{h}$, so if $y = g(x)$ then why wouldn't it be $\frac{d}{dy}f(g(x)) = \lim_{h \to 0} \frac{f(g(x)+h) - f(g(x))}{h}$? – user51819 Sep 09 '16 at 20:55
  • Because you're supposed to swap $x$ with $x+h$ entirely when calculating $\frac{d}{dx}$. – Arthur Sep 09 '16 at 20:57
  • Here you go! http://math.stackexchange.com/a/237689/272831 In other words, use $$\frac{d}{dx}f(x)=\lim_{h\to x}\frac{f(x)-f(h)}{x-h}$$ – Simply Beautiful Art Sep 09 '16 at 20:57
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    This answer explains how easy it is to derive the chain rule informally, and it's also possible to turn this derivation into a rigorous proof (just by keeping track of the error terms): http://math.stackexchange.com/a/725963/40119 – littleO Sep 09 '16 at 21:06
  • I have to run out for now but I'll look over the answers when I get back (I didn't want anyone thinking I wasn't going to accept anything) – user51819 Sep 09 '16 at 21:20

3 Answers3

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Chain rule is a tricky thing to prove; the proof follows the intuition of "cancellation" in $\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}$. In particular, rewrite your limit as $$ \lim_{h\to 0} \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)} \frac{g(x+h) - g(x)}{h} $$ Getting that first term into the desired form from here requires a careful treatment of the definition of a limit and continuity. Essentially, we need to argue that if we set $u(h)=g(x+h)$, then the limit of the first fraction can be rewritten as $$ \lim_{u\to g(x)} \frac{f(u) - f(g(x))}{u-g(x)} $$ Which is to say that the limit will be $f'(g(x))$.

Ben Grossmann
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  • Note that I'm sidestepping a subtlety here; in this approach, we need to separately handle the case in which $g'(x)=0$. – Ben Grossmann Sep 09 '16 at 23:10
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This one takes some thought, but is essentially just applying the definition of the derivative.

Put $h(x) = g(f(x))$.

By the Mean Value Theorem, we have

$f(t) - f(x) = [t-x][f'(x) + u(t)],$ where $u(t) \to 0$ as $t \to x$,

and, setting $y = f(x)$

$g(f(t)) - g(f(x)) = [f(t) - f(x)][g'(f(x)) + v(f(t)]$, where $v(f(t)) \to 0$ as $f(t) \to f(x)$.

Then we have $h(t) - h(x) = g(f(t))- g(f(x)) = [f(t) - f(x)][g'(f(x)) + v(f(t)] = [t-x][f'(x) + u(t)][g'(f(x)) + v(f(t)]$.

Then, letting $t \to x$, we see that $f(t) \to f(x)$, hence $v(f(t)) \to 0$ and $u(t) \to 0$.

Dividing through by $t-x$ and letting $t \to x$, we arrive at $ lim_{t \to x}\frac {h(t) - h(x)}{t-x} = f'(x)g'(f(x))$, as desired.

David Bowman
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We have $$ \begin{align} \frac{f(g(x+h)) - f(g(x))}{h}&=\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h} \end{align} $$ where clearly the last fraction converges to $g'(x)$. Then notice that for $g(x+h)\neq g(x)$ the first fraction is the slope of a secant on the graph of $f$, but it is undefined for $g(x+h)=g(x)$ due to division by zero. By defining $$ \frac{\Delta f}{\Delta g}(h)= \begin{cases} \frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}&\text{if }g(x+h)\neq g(x)\\ \quad\\ f'(g(x))&\text{if }g(x+h)=g(x) \end{cases} $$ it is straightforward to show that $\frac{\Delta f}{\Delta g}(h)$ is a continuous function of $h$ (because $f$ is differentiable at $g(x)$), and that $$ \begin{align} \frac{f(g(x+h)) - f(g(x))}{h}&=\frac{\Delta f}{\Delta g}(h)\cdot\frac{g(x+h)-g(x)}{h} \end{align} $$ clearly holds when $g(x+h)\neq g(x)$, but also when $g(x+h)=g(x)$ since then both sides become zero.

String
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